Question

A hammer of mass M drops from height ‘h’. It strikes a rail placed vertically on the ground and drives it. Into the ground through distance D. Calculate 

1. The average resistance offered by the ground, assuming that the hammer and rail remain stuck together after impact, 

2. The time for which the rail is in motion and 

3. The loss in kinetic energy in input.

Answer 1

The hammer of mass M falls freely under the gravity through a distance. Let v be the speed acquired by the hammer when it strikes the rail obviously.

v =√gh ……… (1)

On impact, the hammer and rail are stuck together. Let v be the speed of the combination after impact. The law of conservation of momentum gives.

Mv = (M + m) v_¹ …….. (2)

Where m is the mass of the rail. Let FG be the average resistance (or resistive force) exerted by the ground. The net upward force F on the hammer rail combination is

F = F_G – (M + m)g.

The combination moves through a distance ‘d’ against this net upward force. Obviously, the work done against the force F in a distance ‘d’ must equal to the kinetic energy the combination had just after impact.

\(\frac{1}{2}\) (M+m) v¹² = F x d

= [F_G – (M + m) g] × d ……… (3)

By equations (1) & (2), we can rewrite equation (3) as

2. Let the hammer rail combination be moving for a time Δt, before coming to rest. If ΔP is the change in momentum, 

(Δ_p/Δ_t) = F_G – (M + m)g 

|Δ_p| = (M + m) v₁ – 0

3. The kinetic energy of the hammer rail mcombination just before the impact is \(\frac{1}{2}\)(M +m) v².

Mv² and after impact, it is

Loss in kinetic energy = \(\frac{1}{2}\)Mv² – \(\frac{1}{2}\).